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guerrero25
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We are given two integers a and b, such that, 2 < a < b and b is not a 01 Apr 2013, 01:05
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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95% (hard)

Question Stats:

42% (02:35) correct 58% (02:29) wrong based on 119 sessions

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We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1?

(1) The least common multiple of a and b is 42

(2) The greatest common factor of a and b is 2
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mau5
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We are given two integers a and b, such that, 2 < a < b and b is not a 01 Apr 2013, 01:54
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guerrero25
We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1?

(1) The least common multiple of a and b is 42

(2) The greatest common factor of a and b is 2

I am struggling in this type of questions ..Any help is greatly appreciated. :)

thanks!
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From F.S 1 , for a=6,b=7, we have the remainder as 1, which is not more than 1. Again, for a=3,b=14, we have a remainder as 2, which is more than 1. Insufficient.

From F.S 2, the integers a and b are of the form 2x and 2y, where x and y are co-primes.

Also, 2y = 2x*q + R, where q is a non-negative integer.

or R = 2(y-qx). As x and y are co-primes, thus y is not equal to qx, for any integral value of q. Thus, (y-qx) will never be zero. And as R is always positive, the value of R will always be more than 1. Sufficient.

Note that as a>2, we can have the first value of a only as 4.

B.
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We are given two integers a and b, such that, 2 < a < b and b is not a Updated on: 30 Oct 2021, 11:27
Originally posted by BrushMyQuant on 01 Apr 2013, 02:03.
Last edited by BrushMyQuant on 30 Oct 2021, 11:27, edited 1 time in total.
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STAT1 is not sufficient as we can have mutiple cases
case 1 a=6, b=7
now remainder when b is divided by a will be 1 which is NOT greater than 1

case 2 a=6, b=14
now reaminder when b is divided by a will be 2 which is greater than 1

So, STAT1 is NOT sufficient

STAT2 : GCD of a and b is 2 means that both a and b are even numbers
And we know that b is not a mutiple of a
so, in any case reamindder when b is divided by a will be more than 1

case1 a=4, b=6
remaidner will be 2 greater than 1

case 2 a=6, b=10
reaminder will be 4 greater than 1

Also, if you notice then the reaminder is a even number greater than or equal to 2
So STAT2 is SUFFICIENT

So, Answer will be B
Hope it helps!

Watch the following video to Learn the Basics of LCM and GCD

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We are given two integers a and b, such that, 2 < a < b and b is not a 01 Apr 2014, 06:26
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Alright, as Hamilton Lin says, let's do it!

Statement 1 tells us that the LCM of (a,b) is 42. Therefore, we have that b=7,a=6 remainder is 1, but if b=14 and a=3 remainder is 2, therefore two different answers–> Insufficient.

Statement 2, says that the GCF of (a,b) is 2. Therefore since both are even and b is not a multiple of a the remainder will always be 2, therefore Sufficient

B is the correct answer

Cheers
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We are given two integers a and b, such that, 2 < a < b and b is not a 21 Mar 2018, 01:19
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Solution:



We are given:

    • ‘\(b\)’ is not a multiple of ‘\(a\)’.

    • Both, ‘\(a\)’ and ‘\(b\)’, are greater than \(2\) and ‘\(b\)’ is greater than ‘\(a\)’.

Statement-1 is “The least common multiple of ‘a’ and ‘b’ is 42”.

\(42= 2*3*7\)

‘\(a\)’ and ‘\(b\)’ can have multiple values based on prime factorization of \(42\).



Since the value of remainder may or may not greater than 1. Hence, Statement 1 alone is not sufficient to answer the question.

Statement-2 is “The greatest common factor of ‘a’ and ‘b’ is 2.”

Thus,
    • \(a=2x\) and \(b=2y\), where \(x\) and \(y\) are co-prime numbers.

Since ‘\(b\)’ is greater than ‘\(a\)’, ‘\(b\)’ on dividing by ‘\(a\)’ can be written as:

    • \(2y= 2x*m + r\), where ‘\(r\)’ is the remainder and ‘\(r\)’ cannot be ‘\(0\)’ as ‘\(b\)’ is not a multiple of ‘\(a\)’.

    • Since \(2y\) is an even number, \((2x*m + r)\) should be an even number.

      o ‘\(r\)’ must be even.

         Thus, the least value of ‘\(r\)’ can be \(2\).

Hence, Statement 2 alone is sufficient to answer the question.


Answer: B
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